Question

(a) Solve the two initial-value problems:

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{y}\mathbf{+}\frac{y}{\mathrm{xlnx}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{e}\mathbf{\right)}\mathbf{=}\mathbf{1}\end{array}$

(b) Show that there are more than$\mathbf{1}\mathbf{.}\mathbf{65}$million digits in the$\mathit{y}$-coordinate of the point of intersection of the two solution curves in part (a).

Short answer

(a) The solution are$\ln \left(y\right)=x$ &$\ln y=x+\ln \left(\ln x\right)-e$respectively.

(b) The point of intersection of the two solution curve is $e^{2\times {10}^6}$.

Step-by-step solution

Definition

A first-order differential equation of the form $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{h}\mathbf{\left(}\mathit{y}\mathbf{\right)}$is said to be separable or to have separable variables.

Integrate(a)

Consider$\frac{dy}{dx}=y,y\left(0\right)=1$

Solve the initial value problem$\frac{dy}{dx}=y,y\left(0\right)=1$

This equation is separable so we have

$\begin{array}{c}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}\\ \int \frac{1}{\mathrm{y}}\mathrm{dy}=\int \mathrm{d}\mathrm{x}\\ \ln \left(\mathrm{y}\right)=\mathrm{x}+\mathrm{c}\end{array}$

Apply initial condition

Plugging in the initial condition$y\left(0\right)=1$ to solve for $c$.

$\begin{array}{c}\ln \left(1\right)=0+c\\ 0=c\end{array}$

So the solution to the initial value problem is $\ln \left(y\right)=x$.

Integrate

Consider$\frac{dy}{dx}=y+\frac{y}{x\ln x},y\left(e\right)=1$

Solve the initial value problem$\frac{dy}{dx}=y+\frac{y}{x\ln x},y\left(e\right)=1$

This equation is separable if we factor $y$first. So we have

$\begin{array}{c}\frac{dy}{dx}=y\left(1+\frac{1}{x\ln x}\right)\\ \int \frac{1}{y}dy=\int 1+\frac{1}{x\ln x}dx\\ \ln y=x+\ln \left(\ln x\right)+c\end{array}$

Apply initial condition

Now plugging in the initial condition,$y\left(e\right)=1$ we get

$\begin{array}{c}\ln \left(1\right)=\mathrm{e}+\ln \left(\mathrm{lne}\right)+\mathrm{c}\\ 0=\mathrm{e}+\ln \left(1\right)+\mathrm{c}\\ -\mathrm{e}=\mathrm{c}\end{array}$

Remember that the functions $\ln x$and $e^x$are inverse functions so $\ln e=1$. So the solution to the initial value problem is $\ln y=x+\ln \left(\ln x\right)-e$.

Point of intersection(b)

Solving for the intersection point between equations (1) and (2), we have

$\begin{array}{c}x=x+\ln \left(\ln x\right)-e\\ e=\ln \left(\ln x\right)\\ e^e=\ln x\\ e^{{ϵ}^e}=x\end{array}$

Plugging thatinto equation (1) yields the value for $y$at the intersection point:

$\begin{array}{l}\ln y=e^{e^{e^x}}\\ y=e^{e^{e^{e^k}}}\end{array}$

The value$e^{e^{\epsilon }}$ is about $2\times {10}^6$. Then, when we evaluate $e^{2\times {10}^6}$, that value will have over$1.65$ million digits.